Biomedical Engineering Reference
In-Depth Information
Table 7.1
Young's modulus, Shear modulus, elastic limit, and strength of some common solids
Compressive and tensile
strength (fracture point)
(10
7
Pa)
Young's modulus
(10
10
Pa)
Shear modulus
(10
10
Pa)
Elastic limit
(10
7
Pa)
Material
Aluminum
7.0
2.5
18
20
Copper
12.0
4.2
20
40
Granite
5.0
20
Steel
20.0
8.4
30
50
Example 7.1
A rod of steel with 1 mm radius and 50 cm length is submitted to a
tension (traction) of 500 N. Find the final length of the rod. Do the same for an
aluminum rod of equal dimension.
Let us begin by calculating the cross-sectional area of the rod, remembering that
1mm
¼
10
3
m.
r
2
¼
3.14
(10
3
m)
2
A
¼ π
(500 N)/(3.14)(10
3
m)
2
10
6
Pa
σ ¼
T
/
A
¼
¼
159.2
10
6
Pa/(20
10
10
Pa)
10
4
ε ¼ Δ
L
/
L
i
¼ σ
/
Y
¼
159.2
¼
8.0
10
4
)
10
4
m
Δ
L
¼
L
i
ε ¼
(0.50 m)(8.0
¼
4.0
¼
0.4 mm that is the elon-
gation of the steel rod. Hence, its final length is
L
f
¼
50.04 cm. When the tensile
stress is removed, the length of the rod returns to 50 cm.
In the case of the aluminum rod its Young's modulus is smaller than that of steel.
Beforehand, we note that the elongation of this rod must be greater than that of the
steel rod, since its Young's modulus is smaller.
10
6
Pa)/(7.0
10
10
Pa)
10
4
ε ¼ σ
/
Y
¼
(159.2
¼
22.7
10
4
)
10
4
m
Δ
L
¼
(0.50 m)(22.7
¼
11.4
¼
1.14 mm.
Hence,
L
f
¼
50.11 cm.
10
5
m
2
. The rod is
stretched to its elastic limit. (a) Find the tension (traction) applied to the rod; (b) find
the strain in the copper rod; (c) determine the tension required to break this rod.
Exercise 7.2
A marble column of 2.0 m height and cross-sectional area of 25 cm
2
supports a mass of 70 tons. Young's modulus for marble is 6
Exercise 7.1
The cross-sectional area of a copper rod is of 5
10
10
Pa and the
10
7
Pa. Find: (a) the radius of
column; (b) the compressive stress applied to the column; (c) the shortening of the
column; (d) the maximum weight that the column supports.
compressive strength at the fracture point is 20
7.4.2 Shear Modulus S
Let us consider a block on a smooth surface. This block is subjected to a tangential
force called the shear force,
T
, applied parallel to its upper surface while its
opposite face is immovable due to the force of static friction
f
s
, of the same
